Question 644424
find the equation of the hyperbola 
*asymptotes y=3/2x and y=-3/2x passing through (4, √117)
**
Given asymptotes show center of hyperbola at (0,0)
Since given point is above positive slope line of asymptote, hyperbola has a vertical transverse axis
Standard form of hyperbola with vertical transverse axis: {{{(y-k)^2/a^2-(x-h)^2/b^2=1}}}, (h,k)=(x,y) coordinates of center, which is (0,0) in this case
slopes of asymptotes with vertical transverse axis=±a/b=±3/2
b=2a/3
equation:{{{y^2/a^2-x^2/b^2=1}}}
{{{y^2/a^2-x^2/(4a^2/9)=1}}}
Using coordinates of given point(4,√117)
{{{117/a^2-9*16/4a^2=1}}}
{{{117/a^2-144/4a^2=1}}}
{{{117/a^2-36/a^2=1}}}
81/a^2=1
a^2=81
a=9
b=2a/3=6
b^2=36
equation:{{{y^2/81-x^2/36=1}}}